Finding Maximum Range of a Projectile: Velocity and Formula

[phoenix20_06]

Homework Statement

Hello,

I have a language problem that affects my Math problem :-) I hope you can assist me with this.

The Maximum range of a projectile is directly proportional to the square of it's velocity. A baseball pitcher can throw a ball at 60 min/h, with a maximum range of 242ft. What would his maximum range be if he could throw the ball at 70 min/h.

I'm not what min/h means. isn't velocity is in m/s or ft/s?

I'm lost at this point.

I would think it's something like m = (v^2)x

but I'm not sure if it's even a right assumption.

thank you for any input

What would be the formula?

Homework Equations

The Attempt at a Solution

 

[berkeman]

Yeah, there is something wrong with the units as stated. AFAIK, "min" is short for minutes, and not much else. "h" can be short for hours, and there's 60 minutes in an hour, so 60 min/h may be a thing, but it sure isn't a velocity. As you say, velocity will have units of distance/time.

So that needs to be clarified at some point, but I think you can solve the problem without knowing what the actual units are for the velocity.

The problem says that the max range is proportional to the square of the velocity. So write that mathematically like this:

D = K V^2

Where K is some proportionality constant. Then you can write two equations for the two different situations (different initial velocities give different max distances):

D_1 = K {V_1}^2

D_2 = K {V_2}^2

Put in the 60 <whatever units> and 242 ft. into the first equation, and put in what you know so far, the 70 <whatever units> into the 2nd equation. Now you want to solve for D2. Do you have some ideas on how you can do that?

 

[phoenix20_06]

Berkeman thank you for your reply.
I believe the solution would be:

242 = K60^2
x = K 70^2

so
x/242 = (70^2/60^2)
x = (4900/3600)* 242 = 398.38m

It was very helpful :-)

 

[berkeman]

Good job. That min/h thing would have thrown me too! It's only because the units don't matter in this problem, that you were able to ignore the goofy units and solve the problem.

 

[HallsofIvy]

I would suspect that "min/h" is a misprint for "mi/h"- mile per hour. But as Berkeman explained, it doesn't affect your answer.